======2020 CCPC 网络赛====== [[http://acm.hdu.edu.cn/contests/contest_show.php?cid=909|比赛链接]] =====A.===== **upsolved by 2sozx** ====题意==== 给定一个开始全白的二维平面，每次操作选择一个矩形将其涂黑，矩形下面紧贴 $x$ 轴，问每次操作过后黑色区域的周长为多少。操作次数 $n \le 2 \times 10^5$ ====题解==== 由于矩形紧贴 $x$ 轴，矩形上下两条边边长可以用线段覆盖来维护，现在考虑左右两条边的边长。易知操作是一个区间取 $\max$ ，每次的和为 $\sum_{i = 1}^{n - 1}|a_i - a_{i + 1}|$ ,维护区间 $a_i,a_{i + 1}$ 其中一个是最小值的个数即可，区间 $\max$ 用吉老师线段树维护即可。 =====B.===== **upsolved by** ====题意==== ====题解==== =====C.===== **solved by 2sozx** ====题意==== 签到题 ====题解==== 签到题 =====D.===== **upsolved by ** ====题意==== ====题解==== =====E.===== **solved by 2sozx Bazoka13 JJLeo** ====题意==== $t$ 组询问，每组询问给出 $n$ 个数，两个人进行游戏，每次每个人可以选择一个数 $x$ ，若存在 $x = p \times q$ 且 $p \not = 1$ 可以将 $x$ 分成 $p$ 个 $q$ ，无法对 $1$ 进行操作， 问先手赢还是后手赢。$t \le 10^4, x\le 10^9, n\le 10$ ====题解==== 将每个数的质因数个数算出来 $p^i$ 与 $p^j$ $i \not = j$ 且 $p \not = 2$ 时算两个，之后用 $Nim$ 游戏的方法算就行。别问为什么 =====F.===== **upsolved by ** ====题意==== ====题解==== =====G.===== **upsolved by** ====题意==== ====题解==== =====H.===== **upsolved by ** ====题意==== ====题解==== =====I.===== **upsolved by ** ====题意==== ====题解==== =====J.===== **upsolved by ** ====题意==== ====题解==== =====K.===== **solved by 2sozx** ====题意==== 给定矩阵 $A,K$，定义矩阵 $C$ 为 $C_{x,y}=\sum_{i=1}^{min(n-x+1,3)}\sum_{j=1}^{min(n-y+1,3)}A_{x+i-1,y+j-1}K_{i,j}$ 定义 $C^{m}(A,K)=C(C^{m-1}(A,K),K)$ 矩阵 $K$ 中元素和为 $1$ 求 $\lim\limits_{m \to \infty} C$ ，矩阵 $K$ 为 $3\times 3$ ====题解==== 从 $C$ 的最后一位开始推很容易得出当且仅当 $K(1,1) = 1$ 时 $\lim\limits_{m \to \infty} C$ 非零，否则 $C = A$ =====L.===== **upsolved by ** ====题意==== ====题解==== =====M.===== **upsolved by ** ====题意==== ====题解==== =====记录===== before:准备视奸全场结果后排没位置了。书接上文：CSK恰了意面，身体不适\\ 0min:开始分题，ZYF冲1010\\ 2min:ZYF AC，MJX冲1003\\ 14min:MJX AC，CSK冲1007\\ 16min:CSK AC，拿了暂时的 rank1，MJX看1011\\ 41min:MJX PE二发后AC，ZYF冲1002\\ 57min:ZYF AC，ZYF,CSK看1005，MJX看1012，自闭开始\\ 240min:换题讨论，顺了起来，MJX 冲1005，ZYF冲1006\\ 263min:MJX AC\\ 264min:ZYF AC\\ after end:MJX ban掉ZYF1012正解，CSK ban掉ZYF正解，结论：要换题看 =====总结=====